# Can the radius be found?

• Feb 18th 2013, 03:01 PM
tmoria
I hope the image is clear enough

http://i1307.photobucket.com/albums/...ps6d507177.jpg

There are 3 segmented circles, the centres are vertically above each other. The vertical distance between each is 'h'.
All coloured triangles are right-triangles. The white and purple triangles are the same.

Given that a, b, c and h are known, is it possible to obtain equations for 'R' and 'd' in terms of these known quantities?

I feel sure that it should be possible

i.e. I known the unmarked side of the red triangle is $\sqrt{R^2-a^2}$ etc

but am struggling to get anywhere with it.

Any help would be greatly appreciated.

Please say if you require any further info.
• Feb 18th 2013, 03:47 PM
ILikeSerena
Re: Can the radius be found?
Hi tmoria! :)

You do have a constraint.
Since the segments a, b, and c are in the same plane and bounded by 2 lines, you get the constraint: a-2b+c=0.

So let's suppose that a and b are known, then we require that:

$c=2b-a \qquad (1)$

Let's call the unnamed line segments A, B, and C.
Then we can set up the system of equations:

$A-2B+C=0$

$A^2+a^2=R^2$

$B^2+b^2=(R+d)^2$

$C^2+c^2=(R+2d)^2$

We have 4 equations with 5 unknowns.

So let's pick A ourselves, then we can solve the system getting:

$R=\sqrt{a^2 + A^2}$

$d=\frac{b-a}{a}R$

$B=\frac b a A$

$C=\frac c a A$

A solution like this is easiest to calculate with Wolfram|Alpha.
• Feb 19th 2013, 12:44 PM
tmoria
Re: Can the radius be found?
Quote:

Originally Posted by ILikeSerena

...Since the segments a, b, and c are in the same plane and bounded by 2 lines..., you get the constraint: a-2b+c=0.

Hi ILikeSerena,

Many thanks for the effort you put in. Unfortunately I forgot to delete the two lines at the end of the segments as they are not straight lines, but slightly curved which does not show in the image. The amended image is included below. I apologize for you wasting your time :(

Lines MN and OP are straight.

http://i1307.photobucket.com/albums/...pse640f131.jpg